Photoelectric Effect: Emitted Photon Frequency

Emitted Photon Frequency in a Photoelectric Process

Question: A free electron of 1.5 eV energy collides with an H⁺ ion. This results in the formation of a hydrogen atom in the second excited state and a photon is released. Given \[ h = 6.6 \times 10^{-34}\, \text{Js}, \] find the frequency of the emitted photon.

(1) \(0.73 \times 10^{9}\) MHz

(2) \(6.2 \times 10^{9}\) MHz

(3) \(0.1 \times 10^{8}\) MHz

(4) \(2 \times 10^{8}\) MHz

Detailed Step-by-Step Explanation

Step 1: Identify the Process and Energy Levels
In this process, a free electron (energy = 1.5 eV) is captured by an H⁺ ion to form a hydrogen atom in the second excited state. For a hydrogen atom, the energy level is given by: \[ E_n = -\frac{13.6\,\text{eV}}{n^2}. \] The second excited state corresponds to \(n = 3\) (since the ground state is \(n = 1\) and first excited is \(n = 2\)). Thus: \[ E_3 = -\frac{13.6}{9}\, \text{eV} \approx -1.51\,\text{eV}. \]

Step 2: Energy Conservation
The free electron is coming in with 1.5 eV (its kinetic energy, relative to the free state defined as 0 eV). After capture, the electron is in an energy state \(E_3 = -1.51\,\text{eV}\). Thus, the energy released as a photon is: \[ E_{\text{photon}} = \left( \text{Initial energy} \right) - \left( \text{Final energy} \right). \] Since the final energy is negative, \[ E_{\text{photon}} = 1.5\,\text{eV} - (-1.51\,\text{eV}) \approx 3.01\,\text{eV}. \]

Step 3: Convert Photon Energy to Joules
Using \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\): \[ E_{\text{photon}} = 3.01 \times 1.6 \times 10^{-19}\,\text{J} \approx 4.82 \times 10^{-19}\,\text{J}. \]

Step 4: Determine Frequency of the Emitted Photon
The energy of a photon is related to its frequency by: \[ E = h \nu. \] Solving for the frequency \(\nu\): \[ \nu = \frac{E}{h} = \frac{4.82 \times 10^{-19}\,\text{J}}{6.6 \times 10^{-34}\,\text{Js}} \approx 7.3 \times 10^{14}\,\text{Hz}. \]

Step 5: Express Frequency in MHz
Note that \(1\,\text{MHz} = 10^6\,\text{Hz}\), so: \[ 7.3 \times 10^{14}\,\text{Hz} = 0.73 \times 10^9\,\text{MHz}. \]

Final Answer: The frequency of the emitted photon is \(0.73 \times 10^{9}\,\text{MHz}\), which corresponds to option (1).

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